(183) Through a wood. When a wood intervenes between

other, as in the figure, in which A and Z are the given points, proceed thus. Hold a rod at some point B' as nearly in the desired line from A as can be guessed at, and as far from A as possible. To approximate to the proper direction, an assistant may be sent to the other end of the line, and his shouts will indicate the direction; or a gun may be fired there; or, if very distant, a rocket may be sent up after dark. Then range out the "random line" AB', by the method given in Art. (169), noting also the distance from A to each point found, till you arrive at a point Z', opposite to the point Z, i. e. at that point of the line from which a perpendicular there erected would strike the point Z. Measure Z'Z. Then move each of the stakes, perpendicularly from the line AZ', a distance proportional to their distances from A. Thus, if AZ' be 1000 links, and ZZ be 10 links, then a stake B', 200 links from A, should be moved 2 links to a point B, which will be in the desired straight line AZ; if C' be 400 links from A, it should be moved 4 links to C, and so with the rest. The line should then be cleared, and the accuracy of the position, of these stakes tested by ranging from A to Z.

(184) To an invisible intersection. Let AB and CD be two

Set stakes at the five given points, A, B, C, D, P. Set a sixth stake at E, in the alinements of AD and CP; and a seventh stake

at F, in the alinements of BC and AP. at G, in the alinements of BE and DF.

line.

Then set an eighth stake
PG will be the required

Otherwise; Through P range out a parallel to the line BD. Note the points where this parallel meets AB and CD, and call these points Q and R. Then the distance from B, on the line BD, to a point which shall be

invisible point, will be =

in the required line running from P to the BDXQP

QR

II. OBSTACLES TO MEASUREMENT.

(185) The cases, in which the direct measurement of a line is prevented by various obstacles, may be reduced to three.

A. When both ends of the line are accessible.

B. When one end of it is inaccessible.

C. When both ends of it are inaccessible.

A. WHEN BOTH ENDS OF THE LINE ARE ACCESSIBLE.

AB; turn a second right angle at B,

B

C

and measure past the obstacle; turn a third right angle at C; and measure to the original line at D. Then will the measured distance, BC, be equal to the desired distance, AD.

If the direction of the line is also unknown, it will be most easily obtained by the additional perpendiculars shown in Fig. 109, of Art. (171).

(188) By symmetrical triangles. Let AB be the distance required. Measure from A obliquely to some A point C, past the obstacle. Measure onward, in the same line, till CD is as long as AC. Place stakes at C and D. From B measure to C, and from C measure onward, in the same line, till CE is equal to CB. be equal to AB, the distance required. CD and CE equal, respectively, to half AB be equal to twice DE.

Let

(189) By transversals. AB be the required distance. Set a stake, C, in the line prolonged; set another stake, D, so that C and B can be seen from it; and a third stake, E, in the line of BD prolonged, and at a distance from D equal to the distance from D to B.

C

Set a fourth stake, F, at the intersection of EA and CD. Measure

AC, AF and FE. Then is AB =

AC

(FE-AF). AF

AB =

= √{(AC — BC)2 + [DE2 — (CE-CD)2] ACBC

CDXCE

As this expression is somewhat complicated, an example will be given: Let AC100, CE=40, CD 30, DE 21, and CB = 80; then will AB 51.7.

=

=

= =

B. WHEN ONE END OF THE LINE IS INACCESSIBLE.

(191) By perpendiculars. This principle may be applied in a variety of ways. In Fig. 125, let AB be the required distance. At the point A, set off AC, perpendicular to AB, and of any convenient length. At C, set off a perpen- c dicular to CB, and continue it to a point, D, in the line of A and B. Measure DA. Then is AC2 AD

AB

=

(192) Otherwise: At the point A, in Fig. 126, set off a perpendicular, AC. At C set off another perpendicular, CD. Find a point, E, in the line of AC, and BD. Measure AE

If EC be made equal to AE, and D be set

Fig. 125.

B

A

in the line of BE, and also in the perpendicular D from C, then will CD be equal to AB.

If EC AE, then CD=AB.

=

(193) Otherwise: At A, in Fig. 127, measure a perpendicular, AC, to the line AB; and at any point, as D, in this line, set off a perpendicular to DB, and continue it to a point E, in the line of CB. Measure DE and also DA. AC X AD = DE AC

Then is AB:

AC, in any convenient direction. From a point D, in the line of BC, measure a line parallel to CA, to a point E, in the line of AB. Measure also AE.

Fig. 129

B

A

(195) By a parallelogram. Set a stake, C, in the line of A and B, and set another stake, D, wherever convenient. With a distance equal to CD, describe from A, an arc on the ground; and, with a distance equal to AC, describe another E are from D, intersecting the first are in E. Or, take AC and CD, so that together they make one chain; fix the ends of the chain at A and D; n take hold of the chain at such a link, that one part of it equals AC, and the other CD, and draw it tight to fix the point E. Set a stake at F, in the intersection of AE and DB.

Measure AF and

AC x CD

EF

the same direction, till CD AC. Take any point E in the line of A and B. Measure from E to C, and onward in the same line, till CF-CE. Then find by trial a point G, which shall be at the same time in the line of C and B, and in G'

the line of D and F. Measure the distance from G to D, and it will be equal to the required distance from A to B. If more convenient, make CD = AC, and CFCE, as shown by the finely dotted lines in the figure. Then will DG

=

AB.